Question: $\overline{AB} = \sqrt{106}$ $\overline{AC} = {?}$ $A$ $C$ $B$ $\sqrt{106}$ $?$ $ \sin( \angle BAC ) = \frac{9\sqrt{106} }{106}, \cos( \angle BAC ) = \frac{5\sqrt{106} }{106}, \tan( \angle BAC ) = \dfrac{9}{5}$
Explanation: $\overline{AB}$ is the hypotenuse $\overline{AC}$ is adjacent to $\angle BAC$ SOH CAH TOA We know the hypotenuse and need to solve for the adjacent side so we can use the cos function (CAH) $ \cos( \angle BAC ) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{\overline{AC}}{\overline{AB}}= \frac{\overline{AC}}{\sqrt{106}} $ $ \overline{AC}=\sqrt{106} \cdot \cos( \angle BAC ) = \sqrt{106} \cdot \frac{5\sqrt{106} }{106} = 5$